Problem: Divide the following complex numbers. $\dfrac{4+19i}{-5-2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-5+2i}$. $ \dfrac{4+19i}{-5-2i} = \dfrac{4+19i}{-5-2i} \cdot \dfrac{{-5+2i}}{{-5+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(4+19i) \cdot (-5+2i)} {(-5)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(4+19i) \cdot (-5+2i)} {(-5)^2 - (-2i)^2} $ $ = \dfrac{(4+19i) \cdot (-5+2i)} {25 + 4} $ $ = \dfrac{(4+19i) \cdot (-5+2i)} {29} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({4+19i}) \cdot ({-5+2i})} {29} $ $ = \dfrac{{4} \cdot {(-5)} + {19} \cdot {(-5) i} + {4} \cdot {2 i} + {19} \cdot {2 i^2}} {29} $ $ = \dfrac{-20 - 95i + 8i + 38 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{-20 - 95i + 8i - 38} {29} = \dfrac{-58 - 87i} {29} = -2-3i $